Q: A beam of electrons enter at right angles to uniform electric field with a velocity 3 x 10^{7} m/s, E = 1800 v/m while travelling through a distance of 10 cm , the beam deflected by 2 mm then specific charge of electron is

Sol: $\large t = \frac{x}{u} = \frac{0.1}{3 \times 10^7} $

$\large t = \frac{1}{3}\times 10^8 sec $

$\large y = \frac{1}{2} a t^2 $

As , a = qE/m ;

$\large y = \frac{1}{2} (\frac{q}{m}E) t^2 $

$\large 2 \times 10^{-3} = \frac{1}{2} (\frac{q}{m}\times 1800) (\frac{1}{3}\times 10^8)^2 $

q/m = 2 × 10^{11} C/ kg.